Calculating Orbital Velocity Change for Rendezvous Problem

[ColdFusion85]

See attached image for problem description and diagram.

I am confused as to how to solve this problem (part A) because of the following. I know the altitude, and hence, radius of the circular orbit that both satellites are initially in. Since we know the radius, we can calculate the period of the circular orbit at r=6778.14 km. Also, we know the velocity via the vis-viva relation. Calculating, I get an orbital period of 5553.631036 seconds and a velocity of 7.668552229 km/s. I left so many significant figures because I don't know how precise I will need to be yet. Now, the satellite T is 4 n.mi or 7.408 km (1 n.mi = 1.852 km) ahead of A. If the two satellites are to dock at point P after A makes one full orbit on its new trajectory, then satellite T will arrive at P in [(2*pi*R)-7.408 km]/[7.668552229 km/s] = 5552.666626 seconds from the time of the maneuver at P to put satellite A on its new trajectory that will ultimately meet with satellite A at point P.

The problem for me is How can we calculate the change in velocity required for satellite A from its original velocity on the circular orbit if we do not have any information about the new orbit's radius, or even if it is circular or not? I just can't seem to figure out how we'd find the new velocity of A. Additionally, we are told that satellite A's velocity needs to decrease to put it in the new orbit. But how do we find this velocity?

 

[ColdFusion85]

Anyone? D.H.?

 

[Janus]

From the period of A's new orbit, you can find its semi-major axis. Just use
$$T = 2 \pi \sqrt{\frac{a^3}{GM}}$$
where a is the semimajor axis, and solve for a.

From the semi-major axis you can find the total energy of A in its new orbit (just assume a mass(m) of 1 for satellite A.) with

$$E = \frac{GMm}{2a}$$

the total energy also can be found by.

$$E = \frac{mv^2}{2}-\frac{GMm}{r}$$
where r is the radial distance of the satellite at a given point of the orbit.

by equating these two expressions for energy, you can solve for v, the velocity of satellite A at point P of its new orbit. (Notice how m cancels out, which is why you can set it to any value you want in this problem.)

The difference between v and the velocity of A in its original orbit is your required velocity change.

 

[D H]

The problem for me is How can we calculate the change in velocity required for satellite A from its original velocity on the circular orbit if we do not have any information about the new orbit's radius, or even if it is circular or not?

You know that its not circular because it was circular at the start. Moreover, you know the new period. You calculated in your original post.

From the period of A's new orbit, you can find its semi-major axis. Just use
$$T = 2 \pi sqrt{\frac{a^3}{GM}}$$
where a is the semimajor axis, and solve for a.

From this point, you could follow the rest of Janus' post, or you could just apply the vis-viva equation.

 

[ColdFusion85]

Excellent. I don't know why I didn't see this earlier. Bringing energy into the picture had crossed my mind, but Janus's post made it much more clear. Thanks guys!

 

[D H]

You don't have to bring energy into the picture explicitly. The vis-viva equation does that for you implicitly. Its also involves less calculation.

BTW, you do know that it is better to compute the semimajor axis using the Earth's standard gravitational parameter $u_{\oplus}$ rather than $GM_{\oplus}$?

 

[ColdFusion85]

Yeah I just realized that about energy, and yes we use mu in class, not the product of GM. I got a delta-V of about -0.001 km/s, a deceleration. Now, if we have to fire the engines to get it back up to the initial velocity when we arrive at P, isn't that just + 0.001 km/s, an acceleration, making total delta-V = 0?

 

[ColdFusion85]

I should be more clear. If delta-V2 has to equal delta-V1, but an acceleration now, so delta-V2=0.001 km/s, isn't total delta V -0.001 + 0.001 = 0?

 

[D H]

No. You add the absolute values. You are using the engines to attain the second delta-v as well as the first. The reason we use delta V is because it serves as a first-order estimate of the amount of fuel needed for a mission.

 

[ColdFusion85]

OK, thank you. I should have realized this.